Video Transcript
Find the limit as 𝑥 tends to ∞ of the positive square root of nine 𝑥 squared minus seven over three 𝑥 plus five.
This limit appears a bit trickier at first due to the square root in the numerator. So what we will try to do is we write the denominator using the square root function so that eventually we can take the square root over the whole quotient. Notice that three 𝑥 plus five equals the square root of three 𝑥 plus five all squared. Now, we have a square root in the denominator. And so, we can take the square root over the whole quotient. Distributing the parentheses in the denominator, we obtain nine 𝑥 squared plus 30𝑥 plus 25.
Now using the property that the limit of the square root of a function is equal to the square root of the limit of the function, we can take the limit inside the square root. Now, we will evaluate the limit inside the square root. In order to do this, we will divide each term in the numerator and denominator by the term with the largest exponent, namely, nine 𝑥 squared. Note that this does not change the limit as we are dividing each term in the numerator and denominator by the same function. Simplifying, we obtain the limit as 𝑥 tends to ∞ of one minus seven over nine 𝑥 squared all divided by one plus 10 over three 𝑥 plus 25 over nine 𝑥 squared.
Now using the property that the limit of a quotient of functions is the quotient of the limits of the functions, we can transfer the limit to the numerator and denominator. Now using the property that the limit of a sum of functions is the sum of the limits of the functions and the property that the limit of a constant multiplied by a function is equal to the constant multiplied by the limit of the function, we can rewrite our limit as the limit as 𝑥 tends to ∞ of one minus seven over nine multiplied by the limit as 𝑥 tends to ∞ of one over 𝑥 squared all divided by the limit as 𝑥 tends to ∞ of one plus 10 over three multiplied by the limit as 𝑥 tends to ∞ of one over 𝑥 plus 25 over nine multiplied by the limit as 𝑥 tends to ∞ of one over 𝑥 squared.
Now the limit of a constant sequence is just that constant. So the limit as 𝑥 tends to ∞ of one is just one. The limit as 𝑥 tends to ∞ of the sequence one over 𝑥 to the power of 𝑛 for a fixed real number 𝑛 greater than or equal to one is zero. This is because as 𝑥 gets larger and larger, so do the terms 𝑥 to the power of 𝑛. If any of the fixed real numbers greater than or equal to one, so the terms one over 𝑥 to the power of 𝑛 get smaller and smaller although always staying positive.
So the limit as 𝑥 tends to ∞ of one over 𝑥 squared equals zero and the limit as 𝑥 tends to ∞ of one over 𝑥 equals zero. So we obtain that our original limit is equal to one. In order to find the limit stated in the question, it just remains to take the positive square root of the number one, which is just one. So our final answer is the number one.
